# Resolve integral with importance sample Monte Carlo

I'm trying to compute the integral $$\int_{a}^{b}(\sin( 1 + x ) + \cos( 1 + x ))e^{-x}\ dx$$

using importance sample Monte Carlo method. The exercise ask to use Cauchy Distribution to resolve the integral. Then $$\frac{1}{n}\sum_{i}^{n}\frac{f(x_i)}{g(x_{i})}$$ is an aproximation to the integral, where $g(x) = \frac{1}{\pi(1+x^2)}$(Cauchy distribution). To generate random numbers in a range $[a, b]$ following Cauchy distribution I used the inverse transformation: $$X = \tan(\arctan(a) + U(\arctan(b)-\arctan(a)))\quad \Rightarrow\quad \ U \sim(0,1)$$

but I did not get success.

$$\int_{a}^{x}{\rm P}\pars{\xi}\,\dd\xi = \int_{0}^{U}\dd U'\quad\imp\quad {\arctan\pars{X} - \arctan\pars{a} \over \arctan\pars{b} - \arctan\pars{a}} =U$$ which leads to $\ds{\arctan\pars{X} =\arctan\pars{a} + \bracks{\arctan\pars{b} - \arctan\pars{a}}U}$: \begin{align} X = \tan\pars{\arctan\pars{a} + \bracks{\arctan\pars{b} - \arctan\pars{a}}} ={a + \tan\pars{\bracks{\arctan\pars{b} - \arctan\pars{a}}U} \over 1 - a\tan\pars{\bracks{\arctan\pars{b} - \arctan\pars{a}}U}}\tag{1} \end{align}
$$\sum_{i = 1}^{N}\fermi\pars{x_{i}} \approx N\int_{a}^{b}{\rm P}\pars{x}\fermi\pars{x}\,\dd x$$
$$\int_{a}^{b}{\fermi\pars{x} \over x^{2} + 1} \approx {\arctan\pars{b} - \arctan\pars{a} \over N}\sum_{i = 1}^{N}\fermi\pars{x_{i}}$$
\begin{align} &\color{#000}{\large% \int_{a}^{b}\bracks{\sin\pars{1 + x} + \cos\pars{1 + x}}\expo{-x}\,\dd x} \\[3mm]&\approx\color{#00f}{\large% {\arctan\pars{b} - \arctan\pars{a} \over N}\times} \\[3mm]&\color{#00f}{\large\sum_{i = 1}^{N} \bracks{\sin\pars{1 + x_{i}} + \cos\pars{1 + x_{i}}}\expo{-x_{i}} \pars{x_{i}^{2} + 1}} \end{align} with $\ds{\braces{x_{i}}}$ generated by $\pars{1}$ and associated $\ds{\braces{U_{i}}}$ generated by an uniform distribution in $\ds{\left[0,1\right)}$.
• thanks very much for the help. For example if I wanted to use the exponential distribution, it would be normalized in [a, b] is given by: $P(x) = \frac{e^{-x + a}}{e^{-b} - e^{-a}}$.Sorry about that more this question is that I'm comparing the distributions – user50098 May 11 '14 at 23:17